Question

what is the new PH of a solution if 10ml of 0.1 M Hcl is added to 150ml of a buffer system that is 0.1M Benzoic acid and 0.1M sodium benzoate ( the Ka of benzoic acid is 6.3X10^-5). Thanks

Answer 1

a) When no NaOH has been added
b) After 25.0 ml of NaOH is added
c) After 50.0 ml of NaOH is added
d) After 75.0 ml of NaOH is added
e) After 100.0 ml of NaOH is added
f) After 300.0 ml of NaOH is added

for this part also you can practice

CH3COOH is a weak acid - calculate [H+] from the Ka equation for question a)
When you add NaOH to the acid you produce a buffer solution Use the Henderson Hasselbalch equation
At the half equivalence point , pH = pKa . pKa for the acid is 4.77 . The half equivalence point is c) So the pH here will be 4.77

I will leave you to rework the problem . If you have not made any progress , or someone else has not done the work for you , I will come back to this problem tomorrow.
Edit next day:
No progress from you and no other answer: So we will do this in full
Question a) You have a weak acid. In order to calculate pH you first have to determine the [H+] in the solution. You do this using the Ka equation:

Ka = [H+] [CH3COO-] / [CH3COOH]
You know that [H+] = [CH3COO-] so for product we write [H+]